Question

A small object is placed \(13.0 \mathrm{~cm}\) from the center of a phonograph turntable. It is observed to remain on the table when it rotates at \(33 \frac{1}{3}\) rev/min but slides off when it rotates at \(45.0\) rev/min. Between what limits must the coefficient of static friction between the object and the surface of the turntable lie?

Short answer

The coefficient of static friction \(μ_s\) must lie between the values calculated in Step 3, which are derived from the speed at which the object remains on the table and the speed at which it slides off.

Step-by-step solution

Calculate the Centripetal Acceleration

The centripetal acceleration \(a_c\) can be calculated using the formula \(a_c = rω^2\), where \(r\) is the radius and ω is the angular velocity in rad/s. Note that 1 rev/min = \(2π/60\) rad/s. The turntable speed is given for two scenarios. Calculate the acceleration for each scenario, namely: when the object doesn't move (rotate at 33 1/3 rev/min) and when it start moving (rotate at 45 rev/min). The radius r = 0.13m as given.

Obtain Centripetal Force

The centripetal force \(F_c\) acting on the object can be calculated using the formula \(F_c = ma_c\), where m is the mass of the object and \(a_c\) is the centripetal acceleration. Note that the problem does not give the mass of the object. However, as we are dealing with friction (which depends on the weight of the object, and therefore the mass and gravity), the mass will cancel out in the final calculation. Therefore, you can use a unit mass to simplify calculations.

Calculate Limits for Coefficient of Static Friction

The frictional force must equal the centripetal force for the object to remain in place. This is given by the equation \(F_s = μ_sF_n\), where \(F_s\) is the frictional force, \(μ_s\) is the coefficient of static friction and \(F_n\) is the normal force. As the small object is not moving vertically, the normal force \(F_n\) is just the weight of the object \(mg\). Thus frictional force which equals centripetal force is \(F_s = μ_smg = F_c = ma_c\). Solving for \(μ_s\), we get \(μ_s = a_c/g\). Calculate this for both given scenarios (when the object stays and when it slides) to get the limits of static friction.

Key concepts

Understanding Static Friction

Static friction is a force that counters the initial motion of an object when it is resting on another surface. It's useful to picture static friction as a kind of invisible anchor keeping an object stationary despite applied forces. Once this force is overcome, the object can start moving, transitioning into kinetic friction.

Considering our exercise, as the phonograph turntable speeds up, static friction is what keeps the object from sliding off immediately. It is only when the force due to the table's rotation – the centripetal force – increases beyond the maximum force of static friction that the object can no longer hold its position.

When we calculate the coefficient of static friction, denoted by \( \text{μ}_s \) , we're finding the ratio that represents how much force is needed to overcome that resistance relative to the object's weight. In the exercise, we know the object stays put at one speed but slips at a higher speed, which means the coefficient of static friction has to be between the forces calculated at these two rotational speeds.

The Role of Centripetal Force

The centripetal force is the required force to keep an object moving in a circular path and is directed towards the center of the circle. It's not a separate force in itself, but rather a term for the net force that causes circular motion. In the case of our turntable, this force is created by the static friction between the object and the table's surface.

This force can be more intense depending on the speed of rotation and the distance from the center – faster speeds or larger radii require more force to maintain the circular motion. If the centripetal force is greater than what static friction can provide, the object will slide off, which is observed at the higher rotational speed in the exercise.

By comparing the forces at different speeds, we can create an inequality for the coefficient of static friction because it must be high enough to provide the necessary centripetal force at lower speeds, but it is not enough at higher speeds. This concept helps set the limits of the coefficient of static friction for the particular situation.

Calculating Angular Velocity

Angular velocity, denoted by \( \text{ω} \) , is a measure of how quickly an object rotates or revolves relative to another point, in this case, the center of the turntable. It's expressed in radians per second (rad/s).

In the step-by-step solution, we converted revolutions per minute (rev/min) into radians per second to compute the centripetal acceleration correctly, which is crucial since this acceleration is a function of both the radius \( r \) and the square of the angular velocity \( \text{ω}^2 \) .

The greater the angular velocity, the greater the centripetal acceleration will be, impacting how much static friction is needed to keep the object from sliding. Therefore, understanding how to calculate and interpret angular velocity is essential in solving problems related to circular motion and directly affects the limits of the coefficient of static friction we are trying to find.